Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 10 - Moments of Inertia - Section 10.7 - Mohr's Circle for M oments of Inertia - Problems - Page 560: 65



Work Step by Step

We can find the product of inertia of the shaded area with respect to the x and y axes as follows: $I_{xy, whole}=\bar{I_{xy}}+A\bar{x}\bar{y}$ $\implies I_{xy, whole}=0+4(6)(3)(2)=144in^4$ Similarly $I_{xy, circle}=\bar{I_{xy}}+A\bar{x}\bar{y}$ $\implies I_{xy, circle}=0+\pi (1)^2(4)(2)=25.14in^4$ Now $I_{xy}=I_{xy, whole}-I_{xy, circle}$ We plug in the known values to obtain: $I_{xy}=144-25.14=119in^4$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.